CN102081685B - Method for distinguishing temperature of submersible motor based on lumped parameter model - Google Patents

Abstract

The invention provides a method for distinguishing the temperature of a submersible motor based on a lumped parameter model, belonging to the technical field of an application base of combination of thermodynamics modeling and motor control. The method is used for solving the problem that acquisition of temperature information for the submersible motor by using a temperature sensor is restricted by subsurface environments. The method comprises the following steps: firstly, collecting the three-phase stator voltage and three-phase stator current of the submersible motor; then, respectively processing the three-phase stator voltage and the three-phase stator current by a signal conditioning circuit, wherein the processed signals as original current input signals of a stator and original voltage input signals of the stator; calculating the gross calorific power u of the submersible motor by an industrial personal computer (IPC) in accordance with the original current input signals of the stator and the original voltage input signals of the stator; and allocating the gross calorific power u to each part of the submersible motor, combining equivalent circuits of the submersible motor, constructing a lumped parameter heat network model and solving the lumped parameter heat network model to obtain a temperature value of each part of the submersible motor. The method is applied to distinguishing the temperature of the submersible motor.

Description

Submersible electric machine with oil temperature identifying approach based on lumped parameter heat supply network network model

Technical field

The present invention relates to a kind of submersible electric machine with oil temperature identifying approach, belong to the application foundation technical field that thermodynamics modeling and Electric Machine Control combine based on lumped parameter heat supply network network model.

Background technology

Submersible electric machine with oil is the work in underground 1.0～3.5km depths generally, is connected with ground power supply through sheathed cable, and down-hole ambient temperature is up to 120 ℃; Motor is when underground work; Will certainly produce certain heat and cause that the motor feels hot, and well liquid can be taken away heat again and make motor radiating when being discharged by centrifugal pump; Do not cause electric machine temperature rise to surpass certain limit, otherwise will burn motor because of temperature rise is too high.For the operation that guarantees that it is safe, efficient, stable, must implement monitoring submersible electric machine with oil downhole temperature and change, take place to avoid accident.

For obtaining the temperature information of submersible electric machine with oil, the most directly method is a mounting temperature sensor on motor, but because the unique HTHP characteristics of oil well, problems such as the installation of temperature sensor, signal transmission are all restricting its application in oil well.

Summary of the invention

The objective of the invention is provides a kind of submersible electric machine with oil temperature identifying approach based on lumped parameter heat supply network network model in order to solve the problem that the temperature information that adopts temperature sensor to obtain submersible electric machine with oil receives the restriction of subsurface environment.

The present invention includes following steps:

Step 1: threephase stator voltage and the threephase stator electric current of gathering said submersible electric machine with oil;

Step 2: through signal conditioning circuit threephase stator voltage and threephase stator electric current are handled respectively, and the signal after will handling is as stator primary current input signal and stator primary voltage input signal;

Step 3: adopt industrial computer to calculate the gross calorific power u of submersible electric machine with oil according to stator primary current input signal and stator primary voltage input signal;

Step 4: in each parts that in said industrial computer, the gross calorific power u of said submersible electric machine with oil is assigned to submersible electric machine with oil according to the physical dimension and the power consumption ratio of each parts in the said submersible electric machine with oil; Simultaneously each parts equivalent electrical circuit of submersible electric machine with oil is combined into the equivalent electrical circuit of whole submersible electric machine with oil according to spatial relation, sets up lumped parameter heat supply network network model according to the equivalent electrical circuit of whole submersible electric machine with oil;

Step 5: find the solution lumped parameter heat supply network network model, obtain the temperature value of each parts of submersible electric machine with oil.

Advantage of the present invention is: the present invention has the characteristics of low cost, high reliability, and it is to realize through detection and corresponding calculating to motor stator voltage, stator current signal to obtaining of parameter of electric machine signal.Be only limited to the parameter identification of surface work motor at present both at home and abroad about the electric machine without sensor parameter detection method; The present invention has realized working in the real-time monitoring of the temperature of submersible electric machine with oil in underground 1.0～3.5km oil well through to the detection of motor stator voltage, current signal and the foundation and the analysis of lumped parameter heat supply network network model.Lumped parameter heat supply network network model is used for solving the complex nonlinear problem that the submersible electric machine with oil temperature is debated knowledge with the combination through electric network of classical thermodynamics, nonequilibrium thermodynamics and kinetic theory.

Lumped parameter heat supply network network model in the thermodynamic analysis of submersible electric machine with oil, has realized that the algorithm of network boom theory of mechanics and power electronics topological analysis intersects with network boom mechanical modeling theory application.Lumped parameter heat supply network network model is as a kind of all critical pieces of submersible electric machine with oil and accurate thermal model of heat conduction mechanism contained; Can be than the virtual condition of accurate description motor; Explicit physical meaning, calculated amount are less relatively, directly approach the thermal behavior of motor; Thermal behavior that can rapid evaluation submersible electric machine with oil operational process is used for the engineering calculation of submersible electric machine with oil temperature.The present invention has realized the soft measurement of submersible electric machine with oil temperature through the temperature of each parts of motor stator current value calculating submersible electric machine with oil of actual measurement.

Description of drawings

Fig. 1 is an on-line monitoring synoptic diagram of the present invention;

Fig. 2 is the equivalent circuit diagram of support;

Fig. 3 is the equivalent circuit diagram of stator yoke;

Fig. 4 is the equivalent circuit diagram of stator tooth;

Fig. 5 is the equivalent circuit diagram of stator winding;

Fig. 6 is the equivalent circuit diagram of air gap;

Fig. 7 is the equivalent circuit diagram of end winding;

Fig. 8 is the equivalent circuit diagram of end air gap;

Fig. 9 is the equivalent circuit diagram of rotor winding;

Figure 10 is the equivalent circuit diagram of rotor yoke;

Figure 11 is the equivalent circuit diagram of axle;

Figure 12 is the equivalent circuit diagram of whole submersible electric machine with oil;

Figure 13 is the simple equivalent circuit figure of Figure 12;

Voltage and the current waveform figure of Figure 14 for submersible electric machine with oil is sampled and obtained;

The stator temperature curve map of Figure 15 for adopting the inventive method to obtain;

Figure 16 is a submersible electric machine with oil surface observed temperature curve map.

Embodiment

Embodiment one: below in conjunction with Fig. 1 to Figure 13 this embodiment is described, this embodiment may further comprise the steps:

Step 1: threephase stator voltage and the threephase stator electric current of gathering said submersible electric machine with oil;

Step 2: through signal conditioning circuit threephase stator voltage and threephase stator electric current are handled respectively, and the signal after will handling is as stator primary current input signal and stator primary voltage input signal;

Step 3: adopt industrial computer to calculate the gross calorific power u of submersible electric machine with oil according to stator primary current input signal and stator primary voltage input signal;

Step 4: in each parts that in said industrial computer, the gross calorific power u of said submersible electric machine with oil is assigned to submersible electric machine with oil according to the physical dimension and the power consumption ratio of each parts in the said submersible electric machine with oil; Simultaneously each parts equivalent electrical circuit of submersible electric machine with oil is combined into the equivalent electrical circuit of whole submersible electric machine with oil according to spatial relation, sets up lumped parameter heat supply network network model according to the equivalent electrical circuit of whole submersible electric machine with oil;

Step 5: find the solution lumped parameter heat supply network network model, obtain the temperature value of each parts of submersible electric machine with oil.

In this embodiment threephase stator voltage and threephase stator electric current are sampled; Can realize through current sensor and voltage sensor; Stator primary current input signal and stator primary voltage input signal after signal conditioning circuit is handled can pass to industrial computer through data collecting card; Detect the stator temperature that obtains submersible electric machine with oil through temperature polling instrument then; And in industrial computer, compare with the final stator temperature that obtains of the inventive method, according to comparative result lumped parameter heat supply network network model is adjusted the correctness of verification model.

Verify the accuracy of the inventive method.

MATLAB coding capable of using calls sampled data in the data capture card.

Said current sensor and voltage sensor are Hall element; Signal conditioning circuit is based on chip AD623; Industrial computer is the data processing platform (DPP) of core; The present invention only need utilize the physical dimension of submersible electric machine with oil and the electric current and voltage data that sampling comes, and finally obtains needed temperature through foundation and analysis to model.

In this embodiment; All obtain the sine-wave current signal of millivolt level after to submersible electric machine with oil stator voltage and current sample by voltage sensor and current sensor; Adopting signal conditioning circuit to be converted into the voltage signal of millivolt level respectively, and further be enlarged into the ac voltage signal of 0-1.5V, is the ac voltage signal of 0-3.3V by 1.8V direct voltage source lifting in the signal conditioning circuit again; Deliver to the data collecting card input port, as original input signal.

To the series of algorithms process that original input signal carried out, all in industrial computer, realize, and to the last temperature estimation of realizing rotor.

Voltage sensor can use the voltage sensor of the LV28-P model of LEM company, and it has outstanding precision, favorable linearity, and low temperature drift, very strong anti-external interference ability, common-mode rejection ratio is very strong, the characteristics of very fast and bandwidth of reaction time.Current sensor can use the current sensor of the KT75A/P model of KEHAI company, and its measurement range is wide, and frequency characteristic is good, and reaction velocity is fast, and overload capacity is strong.

Embodiment two: this embodiment is described below in conjunction with Figure 12 and Figure 13; This embodiment is for to the further specifying of embodiment one, and adopt industrial computer according to the method for the stator primary current input signal and the gross calorific power u of stator primary voltage input signal calculating submersible electric machine with oil to be in the step 3: the gross calorific power u of said submersible electric machine with oil is iron loss W _i, stator copper loss W _sWith copper loss of rotor W _rWith, its expression formula is u=W _i+ W _s+ W _r,

$W_i=[\frac{1}{{\mathrm{cR}}_m}-\frac{R_d}{{\mathrm{cX}}_m(X_m+{2x}_{\mathrm{sc}})}]{V_p}^2-[1-\frac{{\mathrm{cX}}_m}{X_m+2x_{\mathrm{sc}}}]R_d{I}_{\mathrm{<figure-callout\; id="2"\; label="p"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">p</figure-callout>}}^2,$

$W_s=R_d{I}_{\mathrm{<figure-callout\; id="2"\; label="p"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">p</figure-callout>}}^2,$

$W_r=-\frac{R_z}{X_m(X_m+{2x}_{\mathrm{sc}})}{{\mathrm{<figure-callout\; id="2"\; label="V\; p"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">V</figure-callout>}}_p}^2+\frac{{c^{2}X}_m{R}_z}{X_m+2x_{\mathrm{<figure-callout\; id="2"\; label="sc\; I\; p"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">sc</figure-callout>}}}{I}_p^{}{}_2^{},$

R in the formula _dStator resistance, R for submersible electric machine with oil _zRotor resistance, x for submersible electric machine with oil _ScWhole leakage reactance, X for submersible electric machine with oil _mExcitatory reactance, R for submersible electric machine with oil _mExcitatory resistance, V for submersible electric machine with oil _pEvery phase primary voltage input signal values, I for the submersible electric machine with oil stator _pFor every phase primary current input signal values, the c of submersible electric machine with oil stator is constant.

Said c is slightly larger than 1 constant.

In this embodiment based on of the distribution of submersible electric machine with oil equivalent electrical circuit to the loss of electric machine; Confirm the u in the equation; Input as thermodynamical equilibrium equation; The gross calorific power u of motor is made up of the electric energy and the inner winding loss of motor, the iron loss between the rotor mainly on stator tooth and rotor yoke and tooth, i.e. node J among Figure 12 ₃With node J ₈The energy reduction that is used to drive rotor fan is that heat shows as rotor iron loss.Mechanical loss has been left in the basket, but if desired, reduction is on the thermal value of axle.Copper loss between groove and the rotor-end winding, node J among Figure 12 ₄With node J ₆Ratio according to number of conductors is distributed.

In the practice of thermal model, for example dutycycle is calculated or online temperature detection, and the voltage and current that the loss of motor can be through motor terminals calculates with the equivalent electrical circuit of motor, and is shown in Figure 13, wherein stator impedance z ₁=R _d+ jx ₁, x ₁Be stator leakage reactance, rotor impedance z ₂=R _z+ jx ₂, x ₂Be rotor leakage reactance, excitatory impedance z _m=R _m+ jX _m, R _mBe excitatory resistance, X _mBe excitatory reactance, the leakage reactance x that said motor is total _Sc=x ₁+ cx ₂, V is a supply voltage, i is a stator current, i _M0Be exciting curent, i _zBe rotor reduction current value, s is a motor slip ratio.After motor added the three-phase equilibrium electric current, iron loss W _i, stator copper loss W _s, copper loss of rotor W _rJust express with the electric current and voltage of every phase of motor.

Embodiment three: this embodiment is described below in conjunction with Fig. 2 to Figure 11; This embodiment is further specifying embodiment two; Each parts equivalent electrical circuit of submersible electric machine with oil described in the step 4 is to the thermal convection of its each parts and thermal contact resistance is simulated and reduction

4 1): with the thermal contact resistance reduction of support is resistance R ₁And R ₂The resistor network of forming, wherein resistance R ₁And R ₂Be connected on node J ₀And J ₂Between, R ₁And R ₂Tie point be node J ₁, R ₁Obtain for directly measuring,

${\mathrm{<figure-callout\; id="2"\; label="R"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">R</figure-callout>}}_2=\frac{1}{\mathrm{\&pi;}{\mathrm{<figure-callout\; id="1"\; label="h\; c\; Lr"\; filenames="HDA0000037139090000012.png"\; state="\{\{state\}\}">h</figure-callout>}}_c{\mathrm{Lr}}_{}{}_1},$

H in the formula _cBe the contact coefficient of support and iron core, L is a stator length, r ₁Be stator outer diameter;

With the simulation on thermal convection of support is c ₁:

${\mathrm{<figure-callout\; id="1"\; label="c"\; filenames="HDA0000037139090000012.png"\; state="\{\{state\}\}">c</figure-callout>}}_1=M_e{c}_e+\frac{1}{2}{M}_f{c}_f,$

M in the formula _eBe end cap quality, M _fBe support quality, c _eBe end cap specific heat, c _fBe support specific heat; Thermal convection c ₁From node J ₁Flow into this resistor network;

Four or two): with the thermal contact resistance reduction of stator yoke is resistance R ₃, R ₄, R ₅And R ₆The resistor network of forming, wherein resistance R ₅And R ₆Be connected on node J ₁And J _{4 '}Between, resistance R ₅And R ₆Tie point be node J _{2 '}, resistance R ₄Be connected on node J _{2 '}And J ₂Between, resistance R ₃Be connected on node J ₂And J ₇Between,

$R_3=\frac{L}{6\mathrm{\&pi;}{k}_{\mathrm{la}}({\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000037139090000012.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2-{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2)},$

$R_4=\frac{-1}{4\mathrm{\&pi;}{k}_{\mathrm{lr}}\mathrm{Ls}({\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000037139090000012.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2-{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2)}[{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000037139090000012.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2+{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-\frac{{4\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000037139090000012.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2\ln \left(\frac{r_1}{r_2}\right)}{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000037139090000012.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2-{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2}],$

${\mathrm{<figure-callout\; id="5"\; label="R"\; filenames="HDA0000037139090000032.png"\; state="\{\{state\}\}">R</figure-callout>}}_5=\frac{1}{2\mathrm{\&pi;}{k}_{\mathrm{lr}}\mathrm{Ls}}[1-\frac{{2\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2\ln \left(\frac{r_1}{r_2}\right)}{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000037139090000012.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2-{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2}],$

$R_6=\frac{1}{2\mathrm{\&pi;}{k}_{\mathrm{lr}}\mathrm{Ls}}[\frac{{2\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000037139090000012.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2\ln \left(\frac{r_1}{r_2}\right)}{{\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000037139090000012.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2-{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2}-1],$

R in the formula ₂Be the stator tooth radius, s is a stacking factor, k _LaBe axial thermal conductivity, k _LrBe yoke portion thermal conductivity;

With the simulation on thermal convection of stator yoke is c ₂:

$c_2=\frac{c_l{\mathrm{\&rho;}}_l\mathrm{\&pi;Ls}({\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000037139090000012.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2-{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2)}{2},$

C in the formula _lBe yoke portion specific heat, ρ _lBe yoke portion density; Thermal convection c ₂From node J ₂Flow into this resistor network;

Four or three): with the thermal contact resistance reduction of stator tooth is resistance R ₇, R ₈, R ₉, R ₁₀And R ₁₁The resistor network of forming, wherein resistance R ₁₀And R ₁₁Be connected on node J ₂And J ₅Between, resistance R ₁₀And R ₁₁Tie point be node J _{3 '}, resistance R ₉Be connected on node J _{3 '}And J ₃Between, resistance R ₇Be connected on node J ₃With node J ₇Between, resistance R ₈Be connected on node J ₃With node J ₄Between,

${\mathrm{<figure-callout\; id="7"\; label="R"\; filenames="HDA0000037139090000014.png"\; state="\{\{state\}\}">R</figure-callout>}}_7=\frac{{\mathrm{L\&phi;}}_p}{6\mathrm{\&pi;}{k}_{\mathrm{la}}{\mathrm{\&phi;}}_e({\mathrm{<figure-callout\; id="1"\; label="r"\; filenames="HDA0000037139090000012.png"\; state="\{\{state\}\}">r</figure-callout>}}_1^2-{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2)},$

${\mathrm{<figure-callout\; id="8"\; label="R"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">R</figure-callout>}}_8=\frac{\mathrm{\&pi;}{\mathrm{\&phi;}}_e({\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-r_3^2)}{k_{\mathrm{lr}}\mathrm{Ls}{\mathrm{\&phi;}}_p{(r_2-r_3)}^2{\mathrm{<figure-callout\; id="2"\; label="n"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">n</figure-callout>}}^2},$

$R_9=\frac{-{\mathrm{\&phi;}}_p}{4\mathrm{\&pi;}{k}_{\mathrm{lr}}\mathrm{Ls}{\mathrm{\&phi;}}_e({\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-r_3^2)}[{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2+r_3^2-\frac{4{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2{r}_3^2\ln \frac{r_2}{r_3}}{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-r_3^2}],$

$R_{10}=\frac{{\mathrm{\&phi;}}_{\mathrm{<figure-callout\; id="2"\; label="p"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">p</figure-callout>}}}{2\mathrm{\&pi;}{k}_{\mathrm{lr}}\mathrm{Ls}{\mathrm{\&phi;}}_e}[1-\frac{{2r}_3^2\ln \left(\frac{r_2}{r_3}\right)}{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-r_3^2}],$

${\mathrm{<figure-callout\; id="11"\; label="R"\; filenames="HDA0000037139090000014.png"\; state="\{\{state\}\}">R</figure-callout>}}_{11}=\frac{{\mathrm{\&phi;}}_{\mathrm{<figure-callout\; id="2"\; label="p"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">p</figure-callout>}}}{2\mathrm{\&pi;}{k}_{\mathrm{lr}}{\mathrm{Ls\&phi;}}_e}[\frac{{2\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2\ln \left(\frac{r_2}{r_3}\right)}{{\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-r_3^2}-1],$

N is a number of stator slots in the formula, r ₃Be stator tooth internal diameter, φ _pBe stator tooth pitch, φ _eBe equivalent stator sector;

With the simulation on thermal convection of stator tooth is c ₃:

$c_3=\frac{c_l{\mathrm{\&rho;}}_l\mathrm{\&pi;Ls}{\mathrm{\&phi;}}_e({\mathrm{<figure-callout\; id="2"\; label="r"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">r</figure-callout>}}_2^2-r_3^2)}{2{\mathrm{\&phi;}}_p};$

Thermal convection c ₃From node J ₃Flow into this resistor network;

Four or four): with the thermal contact resistance reduction of stator winding is resistance R ₁₂, R ₁₃, R ₁₄And R ₁₅The resistor network of forming, wherein resistance R ₁₄And R ₁₅Be connected on node J _{4 '}And J ₅Between, resistance R ₁₄And R ₁₅Tie point be node J ₄, resistance R ₁₃Be connected on node J ₄And J ₆Between, resistance R ₁₂Be connected on node J ₄And J ₃Between,

$R_{12}=\frac{{2d}_i}{{\mathrm{\&pi;k}}_i{\mathrm{Lr}}_{4}n}+\frac{1}{2\mathrm{\&pi;}{k}_v\mathrm{LFn}},$

$R_{13}=\frac{L}{{6k}_c{A}_{c}n},$

$R_{14}=\frac{{4d}_i}{{\mathrm{\&pi;k}}_i{\mathrm{Lr}}_{4}n}+\frac{1}{{\mathrm{\&pi;k}}_v\mathrm{LFn}},$

$R_{15}=\frac{1}{{\mathrm{\&pi;k}}_v\mathrm{LFn}},$

R in the formula ₄Be stator winding radius, d _iBe insulation thickness, A _cBe the copper cash cross-sectional area, F is transmissibility factor radially, k _cBe copper conductivity, k _iBe line of rabbet joint conductivity, k _vBe the insullac conductivity,

With the simulation on thermal convection of stator winding is c ₄:

$c_4=\frac{c_c{\mathrm{\&rho;}}_c{A}_c\mathrm{Ln}}{2},$

C in the formula _cBe copper specific heat, ρ _cBe copper density; Thermal convection c ₄From node J ₄Flow into this resistor network;

Four or five): with the thermal contact resistance reduction of air gap is resistance R ₁₆, R ₁₇And R ₁₈The resistor network of forming, wherein resistance R ₁₇And R ₁₈Be connected on node J ₄And J _{8 '}Between, resistance R ₁₇And R ₁₈Tie point be node J ₅, resistance R ₁₆Be connected on node J ₅And J ₃Between,

${\mathrm{<figure-callout\; id="16"\; label="R"\; filenames="HDA0000037139090000022.png"\; state="\{\{state\}\}">R</figure-callout>}}_{16}=\frac{{\mathrm{\&phi;}}_p}{\mathrm{\&pi;}{\mathrm{\&phi;}}_e{r}_3{\mathrm{Lh}}_{2r}},$

$R_{17}=\frac{{\mathrm{\&phi;}}_p}{\mathrm{\&pi;}({\mathrm{\&phi;}}_p-{\mathrm{\&phi;}}_e)r_3{\mathrm{Lh}}_{2r}},$

$R_{18}=\frac{1}{\mathrm{\&pi;}{\mathrm{<figure-callout\; id="5"\; label="r"\; filenames="HDA0000037139090000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_5{\mathrm{Lh}}_{2r}},$

R in the formula ₅Be rotor diameter, h _2rBe the air gap convection coefficient;

Four or six): the thermal contact resistance reduction that will hold winding is a resistance R ₁₉, _R20 and R ₂₁The resistor network of forming, wherein resistance R ₂₀And R ₂₁Be connected in parallel on node J ₆And J ₇Between, resistance R ₁₉Be connected on node J ₆And J ₄Between,

$R_{19}=\frac{l_0\mathrm{\&omega;}}{{\mathrm{nA}}_c{k}_c},$

${\mathrm{<figure-callout\; id="20"\; label="R"\; filenames="HDA0000037139090000023.png,HDA0000037139090000062.png"\; state="\{\{state\}\}">R</figure-callout>}}_{20}=\frac{\mathrm{\&omega;}}{16{\mathrm{\&pi;}}^2\mathrm{RF}{k}_v},$

$R_{21}=\frac{{\mathrm{\&omega;r}}_6^2}{8\mathrm{\&pi;}{r}_4^2{l}_0{\mathrm{Fk}}_{v}n},$

R in the formula ₆Be end winding radius, R leaves l for holding winding center and distance of shaft centers ₀For holding the winding outshot long;

With the simulation on thermal convection of end winding is c ₆:

$c_6=\frac{c_c{\mathrm{\&rho;}}_c(1-\mathrm{\&alpha;}){\mathrm{<figure-callout\; id="2"\; label="V\; c"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">V</figure-callout>}}_c}{2\mathrm{\&omega;}},$

α is groove and end winding volume ratio in the formula, and ω is maximum temperature point and medial temperature ratio, V _cCopper volume for whole winding; Thermal convection c ₆From node J ₆Flow into this resistor network;

Four or seven): with the thermal contact resistance reduction of end air gap is resistance R ₂₂, R ₂₃, R ₂₄, R ₂₅, R ₂₆, R ₂₇And R ₂₈The resistor network of forming, resistance R ₂₅And R ₂₆Be connected on node J ₆And J ₈Between, resistance R ₂₅And R ₂₆Tie point be node J ₇, resistance R ₂₂Be connected on node J ₇And J ₁Between, resistance R ₂₃Be connected on node J ₇And J ₂Between, resistance R ₂₄Be connected on node J ₇And J ₃Between, resistance R ₂₇And R ₂₈Be connected in parallel on node J ₇And J ₉Between,

$R_{22}=\frac{1}{A_1{h}_{3r}},$

$R_{23}=\frac{1}{A_2{h}_{3r}},$

$R_{24}=\frac{1}{A_3{h}_{3r}},$

$R_{25}=\frac{1}{A_4{h}_{3r}},$

${\mathrm{<figure-callout\; id="26"\; label="R"\; filenames="HDA0000037139090000024.png"\; state="\{\{state\}\}">R</figure-callout>}}_{26}=\frac{1}{A_5{h}_{3r}},$

$R_{27}=\frac{1}{A_6{h}_{3r}},$

$R_{28}=\frac{1}{A_7{h}_{3r}},$

A in the formula ₁Be end cap contact area, A ₂Be stator yoke contact area, A ₃Be stator tooth contact area, A ₄Be end winding contact area, A ₅Be rotor end ring contact area, A ₆Be rotor yoke contact area, A ₇Be rotor cooling holes contact area, h _3rBe rotating end cap convection coefficient, h _4rBe rotor pore membrane layer coefficient of heat transfer;

Four or eight): with the thermal contact resistance reduction of rotor winding is resistance R ₂₉, R ₃₀, R ₃₁And R ₃₂The resistor network of forming, wherein resistance R ₃₁And R ₃₂Be connected on node J ₅And J ₉Between, resistance R ₃₁And R ₃₂Tie point be node J _{8 '}, resistance R ₃₀Be connected on node J _{8 '}And J ₈Between, resistance R ₂₉Be connected on node J ₈And J ₇Between,

$R_{29}=\frac{L}{6\mathrm{\&pi;}{k}_a({\mathrm{<figure-callout\; id="5"\; label="r"\; filenames="HDA0000037139090000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_5^2-r_6^2)}+\frac{l_e}{\mathrm{\&pi;}{k}_a({\mathrm{<figure-callout\; id="5"\; label="r"\; filenames="HDA0000037139090000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_5^2-{\mathrm{<figure-callout\; id="7"\; label="r"\; filenames="HDA0000037139090000014.png"\; state="\{\{state\}\}">r</figure-callout>}}_7^2)},$

${\mathrm{<figure-callout\; id="30"\; label="R"\; filenames="HDA0000037139090000062.png"\; state="\{\{state\}\}">R</figure-callout>}}_{30}=\frac{-1}{4\mathrm{\&pi;}{k}_{a}L({\mathrm{<figure-callout\; id="5"\; label="r"\; filenames="HDA0000037139090000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_5^2-{\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2)}[{\mathrm{<figure-callout\; id="5"\; label="r"\; filenames="HDA0000037139090000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_5^2+{\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2-\frac{{4\mathrm{<figure-callout\; id="5"\; label="r"\; filenames="HDA0000037139090000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_5^2{\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2\ln \left(\frac{r_5}{r_8}\right)}{{\mathrm{<figure-callout\; id="5"\; label="r"\; filenames="HDA0000037139090000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_5^2-{\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2}],$

${\mathrm{<figure-callout\; id="31"\; label="R"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">R</figure-callout>}}_{31}=\frac{1}{2\mathrm{\&pi;}{k}_{a}L}[1-\frac{{2\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2\ln \left(\frac{r_5}{r_8}\right)}{{\mathrm{<figure-callout\; id="5"\; label="r"\; filenames="HDA0000037139090000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_5^2-{\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2}],$

${\mathrm{<figure-callout\; id="32"\; label="R"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">R</figure-callout>}}_{32}=\frac{1}{2\mathrm{\&pi;}{k}_{a}L}[\frac{{2\mathrm{<figure-callout\; id="5"\; label="r"\; filenames="HDA0000037139090000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_5^2\ln \left(\frac{r_5}{r_8}\right)}{{\mathrm{<figure-callout\; id="5"\; label="r"\; filenames="HDA0000037139090000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_5^2-{\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2}-1],$

R in the formula ₇Be end ring internal diameter, r ₈Be rotor winding radius, l _eBe end ring width, k _aThermal conductivity for aluminium;

With the simulation on thermal convection of rotor winding is c ₈:

$c_8=c_a{\mathrm{\&rho;}}_a[\frac{\mathrm{\&pi;L}({\mathrm{<figure-callout\; id="5"\; label="r"\; filenames="HDA0000037139090000032.png"\; state="\{\{state\}\}">r</figure-callout>}}_5^2-{\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2)}{2}+V_e],$

V in the formula _eBe the volume of end ring and fan, c _aBe the specific heat of aluminium, ρ _aDensity for aluminium; Thermal convection c ₈From node J ₈Flow into this resistor network;

Four or nine): with the thermal contact resistance reduction of rotor yoke is resistance R ₃₃, R ₃₄, R ₃₅And R ₃₆The resistor network of forming, wherein resistance R ₃₅And R ₃₆Be connected on node J ₈And J ₁₀Between, resistance R ₃₅And R ₃₆Tie point be node J _{9 '}, resistance R ₃₄Be connected on node J _{9 '}And J ₉Between, resistance R ₃₃Be connected on node J ₉And J ₇Between,

$R_{33}=\frac{L}{6\mathrm{\&pi;}{k}_{\mathrm{la}}({\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2-r_9^2-{\mathrm{mr}}_{10}^2)},$

$R_{34}=\frac{-1}{4\mathrm{\&pi;}{k}_{\mathrm{lr}}\mathrm{Ls}({\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2-r_9^2-{\mathrm{mr}}_{10}^2)}[r_9^2+{\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2-\frac{{4r}_9^2{\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2\ln \left(\frac{r_8}{r_9}\right)}{{\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2-r_9^2}],$

${\mathrm{<figure-callout\; id="35"\; label="R"\; filenames="HDA0000037139090000032.png"\; state="\{\{state\}\}">R</figure-callout>}}_{35}=\frac{{\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2-r_9^2}{2\mathrm{\&pi;}{k}_{\mathrm{lr}}\mathrm{Ls}({\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2-r_9^2-{\mathrm{mr}}_{10}^2)}[1-\frac{{2r}_9^2\ln \left(\frac{r_8}{r_9}\right)}{{\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2-r_9^2}],$

$R_{36}=\frac{{\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2-r_9^2}{2\mathrm{\&pi;}{k}_{\mathrm{lr}}\mathrm{Ls}({\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2-r_9^2-{\mathrm{mr}}_{10}^2)}[\frac{{2\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2\ln \left(\frac{r_8}{r_9}\right)}{{\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2-r_9^2}-1],$

R in the formula ₉Be the diameter of axle, r ₁₀Be the cooling holes radius, m is the cooling holes number;

With the simulation on thermal convection of rotor yoke is c ₉:

$c_9=\frac{c_l{\mathrm{\&rho;}}_l\mathrm{Ls}({\mathrm{<figure-callout\; id="8"\; label="r"\; filenames="HDA0000037139090000031.png"\; state="\{\{state\}\}">r</figure-callout>}}_8^2-r_9^2-{\mathrm{mr}}_{10}^2)}{2};$

Thermal convection c ₉From node J ₉Flow into this resistor network; 40): with the thermal contact resistance reduction of axle is resistance R ₃₇And R ₃₈The resistor network of forming, resistance R ₃₇And R ₃₈Be connected on node J ₉And J ₁Between, resistance R ₃₇And R ₃₈Tie point be node J ₁₀,

${\mathrm{<figure-callout\; id="37"\; label="R"\; filenames="HDA0000037139090000033.png"\; state="\{\{state\}\}">R</figure-callout>}}_{37}=\frac{1}{2\mathrm{\&pi;}{k}_{s}L}+\frac{l_m}{2\mathrm{\&pi;}{k}_s{r}_9^2},$

${\mathrm{<figure-callout\; id="38"\; label="R"\; filenames="HDA0000037139090000033.png"\; state="\{\{state\}\}">R</figure-callout>}}_{38}=\frac{1}{4\mathrm{\&pi;}{k}_s{l}_b}+\frac{l_m}{2\mathrm{\&pi;}{k}_s{r}_9^2},$

L in the formula _bWide for bearing housing, l _mBe the distance of bearing center to rotor center, k _sConductivity for axle;

With the simulation on thermal convection of axle is c ₁₀:

$c_{10}={\mathrm{\&rho;}}_s{c}_s\mathrm{\&pi;}{r}_9^2(l_m+\frac{1}{2}{l}_b+\frac{1}{6}L),$

C in the formula _sBe the specific heat of axle, ρ _sBe the density of axle, thermal convection c ₁₀From node J ₁₀Flow into this resistor network.

Support component described in this embodiment comprises whole rib shape cooling structures and end cap; When it is carried out equivalence the temperature equalization of postulated mechanism base member total and also only through support with heat transferred to outside, it absorbs heat through the convection action of contact thermal resistance between support and the stator and end cap air gap from stator.

The part that described stator yoke is removed stator tooth by stator core constitutes, and the modeling method of its modeling and the above-described basic element of character is similar.At last make certain modification according to the conductivity of stacking factor and mild carbon steel.

The equivalence of described stator tooth is through the stator tooth equivalence thermo-contact for the circular fragment of series of parallel, comes stator tooth is carried out modeling and calculates the equivalent radian φ of tooth with the basic element of character equivalent electrical circuit of expanding _eProvide the sectional area of actual profile of tooth.The mobile thermal resistance that is used between rooved face and the tooth center medial temperature point of annular heat between the groove winding is simulated.Method for determining dimension is identical with stator yoke.

The winding of said stator winding in groove partly simulated with the equivalent method of the basic element of character, and it is made up of a series of conductor and insulator.When confirming radial and axial conductor, that suppose groove axially has only copper conductor transmission heat, but is homogeneous solid with the radial conductor equivalence, and its volume is 2.5 times of insulator.The selection of the radius in cross section is to be 100% useful area in order to obtain a copper factor.The line of rabbet joint uses the nylon wire that is consistent with the class of insulation of motor to come equivalence.

Said air gap is that stator tooth, stator winding part and the rotor surface in groove provides the passage that transmits heat.Corresponding thermal resistance can calculate according to contact area and air gap film conductance.

Said end winding adopts uniform loop configuration to come equivalence, uses the method same with the stator slot winding.It equals the mean radius of stator slot the external diameter of approximate order end winding.The equivalent redius of annular cross-sectional area can measure by remaining winding after deducting groove and copper conductor wherein.

The equivalence of said end air gap does, supposes the temperature constant of end air gap, only describes the heat conduction between end air gap and all the contacted with it parts with a single film conductance.The area of the end winding that calculates will increase by 50% and remedy the adverse effect that surface imperfection is brought.

The equivalence of said rotor winding does, it is assumed to a circular cylinder around rotor yoke, and its volume equals the volume of cage type aluminum strip.What link to each other with the rotor winding is one and rotor end ring and the isopyknic annular equivalent circular cylinder of fan, comes the heat transmission between equivalent aluminum strip and the rotor core with this simple method.

The equivalent method of said rotor yoke is identical with the equivalent method of stator yoke, if axially cooling holes exists, the thermal resistance of the whole parts of rotor yoke is pressed the proportional increase of reduction of sectional area.

The equivalent method of said axle is: with its equivalence is a right cylinder that does not have thermal source; Axial heat conduction is divided into three parts; Be the part between the medial temperature point of rotor yoke, bearing and these two parts; The medial temperature point of whole axle is the center of these three parts, and axle and support think that good heat transmission is arranged, and any part of stretching out bearing all is considered to the part of support.

Embodiment four: this embodiment is described below in conjunction with Figure 12; This embodiment is further specifying embodiment three; In the said step 4 each parts equivalent electrical circuit of submersible electric machine with oil is combined into the equivalent electrical circuit of whole submersible electric machine with oil according to spatial relation; And a plurality of resistance between a pair of node are carried out equivalence replace, the resistance of equivalence replacement has:

Node J ₁With node J ₁₀Between equivalent resistance R ' ₁₁₀, R ' ₁₁₀=R ₃₈Node J ₁₀With node J _{9 '}Between equivalent resistance R _{9 ' 10}, R _{9 ' 10}=R ₃₆//R ₃₇Node J _{9 '}With node J _{8 '}Between equivalent resistance R _{8 ' 9 '}, R _{8 ' 9 '}=R ₃₂//R ₃₅Node J _{8 '}With node J ₅Between equivalent resistance R _{58 '}, R _{58 '}=R ₁₈//R ₃₁Node J ₅With node J _{3 '}Between equivalent resistance R _{3 ' 5}, R _{3 ' 5}=R ₁₁//R ₁₆Node J _{3 '}With node J _{4 '}Between equivalent resistance R _{3 ' 4 '}, R _{3 ' 4 '}=R ₁₀, node J _{4 '}With node J _{2 '}Between equivalent resistance R _{2 ' 4 '}, R _{2 ' 4 '}=R ₆, node J _{2 '}With node J ₁Between equivalent resistance R _{12 '}, R _{12 '}=R ₂//R ₅

Node J ₁With node J ₇Between equivalent resistance R ' ₇₁, R ' ₇₁=R ₂₂, node J ₇With node J ₂Between equivalent resistance R ' ₂₇, R ' ₂₇=R ₃//R ₂₃, node J ₂With node J _{2 '}Between equivalent resistance R _{22 '}, R _{22 '}=R ₄,

Node J ₇With node J ₃Between equivalent resistance R ' ₃₇, R ' ₃₇=R ₇//R ₂₄, node J ₃With node J _{3 '}Between equivalent resistance R _{33 '}, R _{33 '}=R ₉

Node J ₇With node J ₆Between equivalent resistance R ' ₆₇, R ' ₆₇=R ₂₁//R ₂₀//R ₂₅, node J ₆With node J ₄Between equivalent resistance R ' ₄₆, R ' ₄₆=R ₁₃//R ₁₉, node J ₄With node J _{4 '}Between equivalent resistance R _{44 '}, R _{44 '}=R ₁₄, node J ₄With node J ₃Between equivalent resistance R ' ₃₄, R ' ₃₄=R ₈//R ₁₂, node J ₄Node J ₅Between equivalent resistance R ' ₄₅, R ' ₄₅=R ₁₅//R ₁₇, node J ₇With node J ₈Between equivalent resistance R ' ₇₈, R ' ₇₈=R ₂₆//R ₂₉, node J ₈With node J _{8 '}Between equivalent resistance R _{88 '}, R _{88 '}=R ₃₀, node J ₇With node J ₉Between equivalent resistance R ' ₇₉, R ' ₇₉=R ₂₇//R ₂₈//R ₃₃, node J ₉With node J _{9 '}Between equivalent resistance R _{99 '}, R _{99 '}=R ₃₄

The thermal value of stator winding is u in the above-mentioned equivalent circuit structure ₄, the end winding thermal value be u ₆, stator tooth thermal value be u ₃, the rotor winding thermal value be u ₈, the relational expression between the gross calorific power u of said four thermal values and submersible electric machine with oil is: u=u ₃+ u ₄+ u ₆+ u ₈

The gross calorific power u of submersible electric machine with oil can be assigned in its each parts the input as lumped parameter heat supply network network model according to the power consumption situation of the physical dimension of said submersible electric machine with oil and each parts in this embodiment.

Embodiment five: below in conjunction with Figure 12 this embodiment is described, this embodiment is for to the further specifying of embodiment four, and the lumped parameter heat supply network network model described in the step 4 is:

θ wherein ₁Be support temperature, θ ₂Be stator yoke temperature, θ ₃Be stator tooth temperature, θ ₄Be stator winding temperature, θ ₅Be air gap temperature, θ ₆Be end winding temperature, θ ₇Be end air gap temperature, θ ₈Be rotor winding temperature, θ ₉The rotor yoke temperature, θ ₁₀Be the axle temperature degree.

The heat supply network of lumped parameter described in this embodiment network model adopts the method for analogy electric system that the heat supply network network is carried out equivalence, and utilizes the Kirchhoff's second law row to write the thermodynamical equilibrium equation of each node temperature in the circuit shown in Figure 12.The theoretical network that adopts electronics and electric system of its heat supply network network; Make full use of electric aspect and find the solution data and the experience that various complex nonlinear problems are accumulated; Introduce two laws of conservation of kirchhoff and the Tellegen that extension obtains thereof; Be converted into the electrical network of finding the solution non-electric problem, after the thermodynamic (al) discretize of network, become the problem that the general matrix method is found the solution easily.

Shown in Figure 12, described lumped parameter heat supply network network model has 15 nodes, writes differential equation group according to nodal method of analysis row and obtains lumped parameter heat supply network network model.

In the said lumped parameter heat supply network network model 8 differential equations and 7 linear equations are arranged; For answering conveniently; Need not contain 7 elimination of unknowns of differential term; Therefore utilize wherein 7 linear equations will not contain the unknown number of differential term counter separate out in all the other 8 differential equations of substitution answer, finally obtain containing the differential equation group of 8 unknown numbers.

Embodiment six: this embodiment is further specifying embodiment five; The method that said step 5 is found the solution lumped parameter heat supply network network model is: adopt the classical Runge-Kutta method of quadravalence to find the solution lumped parameter heat supply network network model, the primary iteration process formula of the classical Runge-Kutta method of said quadravalence is:

$\left\{\begin{array}{c}{y}_{n+1}=y_n+\frac{h}{6}({\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="HDA0000037139090000012.png"\; state="\{\{state\}\}">K</figure-callout>}}_1+{2\mathrm{<figure-callout\; id="2"\; label="K"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">K</figure-callout>}}_2+{2K}_3+K_4)\\ {\mathrm{<figure-callout\; id="1"\; label="K"\; filenames="HDA0000037139090000012.png"\; state="\{\{state\}\}">K</figure-callout>}}_1=f(x_n,y_n)\\ {\mathrm{<figure-callout\; id="2"\; label="K"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">K</figure-callout>}}_2=f(x_n+\frac{\mathrm{<figure-callout\; id="2"\; label="h"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">h</figure-callout>}}{2},y_n+\frac{h}{2}{K}_1)\\ K_3=f(x_n+\frac{\mathrm{<figure-callout\; id="2"\; label="h"\; filenames="HDA0000037139090000012.png,HDA0000037139090000041.png"\; state="\{\{state\}\}">h</figure-callout>}}{2},y_n+\frac{h}{2}{K}_2)\\ K_4=f(x_n+h,y_n+h{K}_3)\end{array}\right.,$

Y in the formula _N+1Be the functional value of the n+1 time iteration, y _nBe the functional value of the n time iteration, K ₁, K ₂, K ₃, K ₄Be respectively intermediate variable, h is an iteration step length.

Embodiment seven: below in conjunction with Figure 14 to Figure 16 this embodiment is described, this embodiment is for to the further specifying of embodiment six, and the expression formula that obtains the temperature value of each parts of submersible electric machine with oil in the said step 5 is:

$\left[\begin{array}{c}{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_1\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_2\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_3\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_4\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_5\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_6\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_7\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_8\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_9\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_{10}\end{array}\right]=\left[\begin{array}{cccccccc}-0.576& 0.6& 0.07& 0.04& 0& 0& 0& 0.01\\ 0.41& 0.64& 0.14& 0.08& 0.01& 0& 0& 0\\ -0.22& 0.62& -2.85& 1.87& 0.01& -2.28& 4.23& -1.37\\ -15.6& 43.8& 95.83& -123.88& 0.36& 0& 0& 0\\ 0& 0.04& 0.01& 0& -0.19& 0.123& 0& 0\\ 0.07& 0.2& 0.56& 0.39& 0.07& 15.55& -24.4& 7.93\\ 0& 0& 0& 0& 0.04& 233.62& -238.69& 89.67\\ 0& 0.006& 0& 0& 0& 3.64& -4.49& 0.85\end{array}\right]\left[\begin{array}{c}{\mathrm{\&theta;}}_1\\ {\mathrm{\&theta;}}_2\\ {\mathrm{\&theta;}}_3\\ {\mathrm{\&theta;}}_4\\ {\mathrm{\&theta;}}_5\\ {\mathrm{\&theta;}}_6\\ {\mathrm{\&theta;}}_7\\ {\mathrm{\&theta;}}_8\\ {\mathrm{\&theta;}}_9\\ {\mathrm{\&theta;}}_{10}\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ 0.052\\ 19.93\\ 0.011\\ 0.077\\ 0\\ \text{0}\end{array}\right]$

The present invention is according to working in submersible electric machine with oil sealing and the specific (special) requirements of insulating material to high temperature, overvoltage in underground 1.0～3.5km oil well; Provide a kind of base area motor stator voltage that surface sample obtains, stator current parameter to come the soft-measuring technique of the underground 1.0～3.5km of real-time identification depths submersible electric machine with oil temperature; Solved the restriction that problems such as installation when utilizing the temperature sensor measurement motor temperature, signal transmission receive, had that the temperature identification effect is good, precision is high, can realize the advantage to the submersible electric machine with oil real time temperature monitoring.The answer of the differential equation group that the core work of software of the present invention is to obtain to the processing of the data of coming through data collecting card sampling and by lumped parameter model.

Realize the real-time monitoring system of the inventive method, form, set its SF, can realize the real-time monitoring of submersible electric machine with oil temperature greater than 5kHz by voltage sensor, current sensor, signal conditioning circuit, data collecting card and industrial computer.

The present invention is based on each elephant exploitation later stage degree of depth of China recovers the oil to the influence of the negative response of submersible electric machine with oil high temperature;, lower-cost advantage high in conjunction with no sensor temperature identification technique identification precision; Through foundation, realize the real time on-line monitoring of submersible electric machine with oil temperature to the accurate thermodynamical model of submersible electric machine with oil; Need not optional equipment to be installed with respect to existing temperature measuring equipments such as polynary testers, have great performance and cost advantage at motor body.

The inventive method is carried out mathematical modeling to submersible electric machine with oil under HTHP; Provide a kind of submersible electric machine with oil not have the important channel of sensor temperature identification; To the real-time control that realizes the submersible electric machine with oil working temperature, guarantee its safe and stable operation and solve the negative effect problem of each elephant exploitation later stage HTHP of present China to have extremely important realistic meaning.

The present invention can satisfy each elephant exploitation later stage of China; In the 2km-3km oil well that influenced by HTHP and cause, power is the needs of the following submersible electric machine with oil running state real-time monitoring of 12kW; And then a situation arises to predict fault; It has solved the cost an arm and a leg problem of increase of the well recovery investment of devices brought and cost for oil production of import polynary tester; Reduced the complexity of on-the-spot oil production equipment, also do not influenced measuring accuracy simultaneously, there is the big deficiency of error in the method that has remedied again based on indirect measurement rotating speeds such as borehole wall vibration analysiss.This ensures that to prolonging the pump detection period and the serviceable life of submersible electric machine with oil it is efficient, safe and stable operation has the extremely using value and the DEVELOPMENT PROSPECT of reality.

Claims (7)

1. submersible electric machine with oil temperature identifying approach based on lumped parameter heat supply network network model, it is characterized in that: it may further comprise the steps:

Step 1: threephase stator voltage and the threephase stator electric current of gathering said submersible electric machine with oil;

Step 2: through signal conditioning circuit threephase stator voltage and threephase stator electric current are handled respectively, and the signal after will handling is as stator primary current input signal and stator primary voltage input signal;

Step 3: adopt industrial computer to calculate the gross calorific power u of submersible electric machine with oil according to stator primary current input signal and stator primary voltage input signal;

Step 4: in each parts that in said industrial computer, the gross calorific power u of said submersible electric machine with oil is assigned to submersible electric machine with oil according to the physical dimension and the power consumption ratio of each parts in the said submersible electric machine with oil; Simultaneously each parts equivalent electrical circuit of submersible electric machine with oil is combined into the equivalent electrical circuit of whole submersible electric machine with oil according to spatial relation, sets up lumped parameter heat supply network network model according to the equivalent electrical circuit of whole submersible electric machine with oil;

Step 5: find the solution lumped parameter heat supply network network model, obtain the temperature value of each parts of submersible electric machine with oil.

2. the submersible electric machine with oil temperature identifying approach based on lumped parameter heat supply network network model according to claim 1 is characterized in that: adopt industrial computer according to the method for the gross calorific power u of stator primary current input signal and stator primary voltage input signal calculating submersible electric machine with oil to be in the step 3: the gross calorific power u of said submersible electric machine with oil is iron loss W _i, stator copper loss W _sWith copper loss of rotor W _rWith, its expression formula is u=W _i+ W _s+ W _r,

$W_i=[\frac{1}{{\mathrm{cR}}_m}-\frac{R_d}{{\mathrm{cX}}_m(X_m+{2x}_{\mathrm{sc}})}]{V_p}^2-[1-\frac{{\mathrm{cX}}_m}{X_m+2x_{\mathrm{sc}}}]R_d{I}_p^2,$

$W_s=R_d{I}_p^2,$

$W_r=-\frac{R_z}{X_m(X_m+{2x}_{\mathrm{sc}})}{V_p}^2+\frac{{c^{2}X}_m{R}_z}{X_m+2x_{\mathrm{sc}}}{I}_p^2,$

R in the formula _dStator resistance, R for submersible electric machine with oil _zRotor resistance, x for submersible electric machine with oil _ScWhole leakage reactance, X for submersible electric machine with oil _mExcitatory reactance, R for submersible electric machine with oil _mExcitatory resistance, V for submersible electric machine with oil _pEvery phase primary voltage input signal values, I for the submersible electric machine with oil stator _pFor every phase primary current input signal values, the c of submersible electric machine with oil stator is constant.

3. the submersible electric machine with oil temperature identifying approach based on lumped parameter heat supply network network model according to claim 2; It is characterized in that: each parts equivalent electrical circuit of submersible electric machine with oil described in the step 4 is to the thermal convection of its each parts and thermal contact resistance is simulated and reduction

4 1): with the thermal contact resistance reduction of support is resistance R ₁And R ₂The resistor network of forming, wherein resistance R ₁And R ₂Be connected on node J ₀And J ₂Between, R ₁And R ₂Tie point be node J ₁, R ₁Obtain for directly measuring,

$R_2=\frac{1}{\mathrm{\&pi;}{h}_c{\mathrm{Lr}}_1},$

H in the formula _cBe the contact coefficient of support and iron core, L is a stator length, r ₁Be stator outer diameter;

With the simulation on thermal convection of support is c ₁:

$c_1=M_e{c}_e+\frac{1}{2}{M}_f{c}_f,$

M in the formula _eBe end cap quality, M _fBe support quality, c _eBe end cap specific heat, c _fBe support specific heat; Thermal convection c ₁From node J ₁Flow into this resistor network;

Four or two): with the thermal contact resistance reduction of stator yoke is resistance R ₃, R ₄, R ₅And R ₆The resistor network of forming, wherein resistance R ₅And R ₆Be connected on node J ₁And J _{4 '}Between, resistance R ₅And R ₆Tie point be node J _{2 '}, resistance R ₄Be connected on node J _{2 '}And J ₂Between, resistance R ₃Be connected on node J ₂And J ₇Between,

$R_3=\frac{L}{6\mathrm{\&pi;}{k}_{\mathrm{la}}(r_1^2-r_2^2)},$

$R_4=\frac{-1}{4\mathrm{\&pi;}{k}_{\mathrm{lr}}\mathrm{Ls}(r_1^2-r_2^2)}[r_1^2+r_2^2-\frac{{4r}_1^2{r}_2^2\ln \left(\frac{r_1}{r_2}\right)}{r_1^2-r_2^2}],$

$R_5=\frac{1}{2\mathrm{\&pi;}{k}_{\mathrm{lr}}\mathrm{Ls}}[1-\frac{{2r}_2^2\ln \left(\frac{r_1}{r_2}\right)}{r_1^2-r_2^2}],$

$R_6=\frac{1}{2\mathrm{\&pi;}{k}_{\mathrm{lr}}\mathrm{Ls}}[\frac{{2r}_1^2\ln \left(\frac{r_1}{r_2}\right)}{r_1^2-r_2^2}-1],$

R in the formula ₂Be the stator tooth radius, s is a stacking factor, k _LaBe axial thermal conductivity, k _LrBe yoke portion thermal conductivity; With the simulation on thermal convection of stator yoke is c ₂:

$c_2=\frac{c_l{\mathrm{\&rho;}}_l\mathrm{\&pi;Ls}(r_1^2-r_2^2)}{2},$

C in the formula _lBe yoke portion specific heat, ρ _lBe yoke portion density; Thermal convection c ₂From node J ₂Flow into this resistor network;

Four or three): with the thermal contact resistance reduction of stator tooth is resistance R ₇, R ₈, R ₉, R ₁₀And R ₁₁The resistor network of forming, wherein resistance R ₁₀And R ₁₁Be connected on node J ₂And J ₅Between, resistance R ₁₀And R ₁₁Tie point be node J _{3 '}, resistance R ₉Be connected on node J _{3 '}And J ₃Between, resistance R ₇Be connected on node J ₃With node J ₇Between, resistance R ₈Be connected on node J ₃With node J ₄Between,

$R_7=\frac{{\mathrm{L\&phi;}}_p}{6\mathrm{\&pi;}{k}_{\mathrm{la}}{\mathrm{\&phi;}}_e(r_1^2-r_2^2)},$

$R_8=\frac{\mathrm{\&pi;}{\mathrm{\&phi;}}_e(r_2^2-r_3^2)}{k_{\mathrm{lr}}\mathrm{Ls}{\mathrm{\&phi;}}_p{(r_2-r_3)}^2{n}^2},$

$R_9=\frac{-{\mathrm{\&phi;}}_p}{4\mathrm{\&pi;}{k}_{\mathrm{lr}}\mathrm{Ls}{\mathrm{\&phi;}}_e(r_2^2-r_3^2)}[r_2^2+r_3^2-\frac{4r_2^2{r}_3^2\ln \frac{r_2}{r_3}}{r_2^2-r_3^2}],$

$R_{10}=\frac{{\mathrm{\&phi;}}_p}{2\mathrm{\&pi;}{k}_{\mathrm{lr}}\mathrm{Ls}{\mathrm{\&phi;}}_e}[1-\frac{{2r}_3^2\ln \left(\frac{r_2}{r_3}\right)}{r_2^2-r_3^2}],$

$R_{11}=\frac{{\mathrm{\&phi;}}_p}{2\mathrm{\&pi;}{k}_{\mathrm{lr}}{\mathrm{Ls\&phi;}}_e}[\frac{{2r}_2^2\ln \left(\frac{r_2}{r_3}\right)}{r_2^2-r_3^2}-1],$

N is a number of stator slots in the formula, r ₃Be stator tooth internal diameter, φ _pBe stator tooth pitch, φ _eBe equivalent stator sector;

With the simulation on thermal convection of stator tooth is c ₃:

$c_3=\frac{c_l{\mathrm{\&rho;}}_l\mathrm{\&pi;Ls}{\mathrm{\&phi;}}_e(r_2^2-r_3^2)}{2{\mathrm{\&phi;}}_p};$

Thermal convection c ₃From node J ₃Flow into this resistor network;

Four or four): with the thermal contact resistance reduction of stator winding is resistance R ₁₂, R ₁₃, R ₁₄And R ₁₅The resistor network of forming, wherein resistance R ₁₄And R ₁₅Be connected on node J _{4 '}And J ₅Between, resistance R ₁₄And R ₁₅Tie point be node J ₄, resistance R ₁₃Be connected on node J ₄And J ₆Between, resistance R ₁₂Be connected on node J ₄And J ₃Between,

$R_{12}=\frac{{2d}_i}{{\mathrm{\&pi;k}}_i{\mathrm{Lr}}_{4}n}+\frac{1}{2\mathrm{\&pi;}{k}_v\mathrm{LFn}},$

$R_{13}=\frac{L}{{6k}_c{A}_{c}n},$

$R_{14}=\frac{{4d}_i}{{\mathrm{\&pi;k}}_i{\mathrm{Lr}}_{4}n}+\frac{1}{{\mathrm{\&pi;k}}_v\mathrm{LFn}},$

$R_{15}=\frac{1}{{\mathrm{\&pi;k}}_v\mathrm{LFn}},$

R in the formula ₄Be stator winding radius, d _iBe insulation thickness, A _cBe the copper cash cross-sectional area, F is transmissibility factor radially, k _cBe copper conductivity, k _iBe line of rabbet joint conductivity, k _vBe the insullac conductivity,

With the simulation on thermal convection of stator winding is c ₄:

$c_4=\frac{c_c{\mathrm{\&rho;}}_c{A}_c\mathrm{Ln}}{2},$

C in the formula _cBe copper specific heat, ρ _cBe copper density; Thermal convection c ₄From node J ₄Flow into this resistor network;

Four or five): with the thermal contact resistance reduction of air gap is resistance R ₁₆, R ₁₇And R ₁₈The resistor network of forming, wherein resistance R ₁₇And R ₁₈Be connected on node J ₄And J _{8 '}Between, resistance R ₁₇And R ₁₈Tie point be node J ₅, resistance R ₁₆Be connected on node J ₅And J ₃Between,

$R_{16}=\frac{{\mathrm{\&phi;}}_p}{\mathrm{\&pi;}{\mathrm{\&phi;}}_e{r}_3{\mathrm{Lh}}_{2r}},$

$R_{17}=\frac{{\mathrm{\&phi;}}_p}{\mathrm{\&pi;}({\mathrm{\&phi;}}_p-{\mathrm{\&phi;}}_e)r_3{\mathrm{Lh}}_{2r}},$

$R_{18}=\frac{1}{\mathrm{\&pi;}{r}_5{\mathrm{Lh}}_{2r}},$

R in the formula ₅Be rotor diameter, h _2rBe the air gap convection coefficient;

Four or six): the thermal contact resistance reduction that will hold winding is a resistance R ₁₉, R ₂₀And R ₂₁The resistor network of forming, wherein resistance R ₂₀And R ₂₁Be connected in parallel on node J ₆And J ₇Between, resistance R ₁₉Be connected on node J ₆And J ₄Between,

$R_{19}=\frac{l_0\mathrm{\&omega;}}{{\mathrm{nA}}_c{k}_c},$

$R_{20}=\frac{\mathrm{\&omega;}}{16{\mathrm{\&pi;}}^2\mathrm{RF}{k}_v},$

$R_{21}=\frac{{\mathrm{\&omega;r}}_6^2}{8\mathrm{\&pi;}{r}_4^2{l}_0{\mathrm{Fk}}_{v}n},$

R in the formula ₆Be end winding radius, R leaves l for holding winding center and distance of shaft centers ₀For holding the winding outshot long;

With the simulation on thermal convection of end winding is c ₆:

$c_6=\frac{c_c{\mathrm{\&rho;}}_c(1-\mathrm{\&alpha;})V_c}{2\mathrm{\&omega;}},$

α is groove and end winding volume ratio in the formula, and ω is maximum temperature point and medial temperature ratio, V _cCopper volume for whole winding; Thermal convection c ₆From node J ₆Flow into this resistor network;

Four or seven): with the thermal contact resistance reduction of end air gap is resistance R ₂₂, R ₂₃, R ₂₄, R ₂₅, R ₂₆, R ₂₇And R ₂₈The resistor network of forming, resistance R ₂₅And R ₂₆Be connected on node J ₆And J ₈Between, resistance R ₂₅And R ₂₆Tie point be node J ₇, resistance R ₂₂Be connected on node J ₇And J ₁Between, resistance R ₂₃Be connected on node J ₇And J ₂Between, resistance R ₂₄Be connected on node J ₇And J ₃Between, resistance R ₂₇And R ₂₈Be connected in parallel on node J ₇And J ₉Between,

$R_{22}=\frac{1}{A_1{h}_{3r}},$

$R_{23}=\frac{1}{A_2{h}_{3r}},$

$R_{24}=\frac{1}{A_3{h}_{3r}},$

$R_{25}=\frac{1}{A_4{h}_{3r}},$

$R_{26}=\frac{1}{A_5{h}_{3r}},$

$R_{27}=\frac{1}{A_6{h}_{3r}},$

$R_{28}=\frac{1}{A_7{h}_{3r}},$

A in the formula ₁Be end cap contact area, A ₂Be stator yoke contact area, A ₃Be stator tooth contact area, A ₄Be end winding contact area, A ₅Be rotor end ring contact area, A ₆Be rotor yoke contact area, A ₇Be rotor cooling holes contact area, h _3rBe the rotating end cap convection coefficient;

Four or eight): with the thermal contact resistance reduction of rotor winding is resistance R ₂₉, R ₃₀, R ₃₁And R ₃₂The resistor network of forming, wherein resistance R ₃₁And R ₃₂Be connected on node J ₅And J ₉Between, resistance R ₃₁And R ₃₂Tie point be node J _{8 '}, resistance R ₃₀Be connected on node J _{8 '}And J ₈Between, resistance R ₂₉Be connected on node J ₈And J ₇Between,

$R_{29}=\frac{L}{6\mathrm{\&pi;}{k}_a(r_5^2-r_6^2)}+\frac{l_e}{\mathrm{\&pi;}{k}_a(r_5^2-r_7^2)},$

$R_{30}=\frac{-1}{4\mathrm{\&pi;}{k}_{a}L(r_5^2-r_8^2)}[r_5^2+r_8^2-\frac{{4r}_5^2{r}_8^2\ln \left(\frac{r_5}{r_8}\right)}{r_5^2-r_8^2}],$

$R_{31}=\frac{1}{2\mathrm{\&pi;}{k}_{a}L}[1-\frac{{2r}_8^2\ln \left(\frac{r_5}{r_8}\right)}{r_5^2-r_8^2}],$

$R_{32}=\frac{1}{2\mathrm{\&pi;}{k}_{a}L}[\frac{{2r}_5^2\ln \left(\frac{r_5}{r_8}\right)}{r_5^2-r_8^2}-1],$

R in the formula ₇Be end ring internal diameter, r ₈Be rotor winding radius, l _eBe end ring width, k _aThermal conductivity for aluminium;

With the simulation on thermal convection of rotor winding is c ₈:

$c_8=c_a{\mathrm{\&rho;}}_a[\frac{\mathrm{\&pi;L}(r_5^2-r_8^2)}{2}+V_e],$

V in the formula _eBe the volume of end ring and fan, c _aBe the specific heat of aluminium, ρ _aDensity for aluminium; Thermal convection c ₈From node J ₈Flow into this resistor network;

Four or nine): with the thermal contact resistance reduction of rotor yoke is resistance R ₃₃, R ₃₄, R ₃₅And R ₃₆The resistor network of forming, wherein resistance R ₃₅And R ₃₆Be connected on node J ₈And J ₁₀Between, resistance R ₃₅And R ₃₆Tie point be node J _{9 '}, resistance R ₃₄Be connected on node J _{9 '}And J ₉Between, resistance R ₃₃Be connected on node J ₉And J ₇Between,

$R_{33}=\frac{L}{6\mathrm{\&pi;}{k}_{\mathrm{la}}(r_8^2-r_9^2-{\mathrm{mr}}_{10}^2)},$

$R_{34}=\frac{-1}{4\mathrm{\&pi;}{k}_{\mathrm{lr}}\mathrm{Ls}(r_8^2-r_9^2-{\mathrm{mr}}_{10}^2)}[r_9^2+r_8^2-\frac{{4r}_9^2{r}_8^2\ln \left(\frac{r_8}{r_9}\right)}{r_8^2-r_9^2}],$

$R_{35}=\frac{r_8^2-r_9^2}{2\mathrm{\&pi;}{k}_{\mathrm{lr}}\mathrm{Ls}(r_8^2-r_9^2-{\mathrm{mr}}_{10}^2)}[1-\frac{{2r}_9^2\ln \left(\frac{r_8}{r_9}\right)}{r_8^2-r_9^2}],$

$R_{36}=\frac{r_8^2-r_9^2}{2\mathrm{\&pi;}{k}_{\mathrm{lr}}\mathrm{Ls}(r_8^2-r_9^2-{\mathrm{mr}}_{10}^2)}[\frac{{2r}_8^2\ln \left(\frac{r_8}{r_9}\right)}{r_8^2-r_9^2}-1],$

R in the formula ₉Be the diameter of axle, r ₁₀Be the cooling holes radius, m is the cooling holes number;

With the simulation on thermal convection of rotor yoke is c ₉:

$c_9=\frac{c_l{\mathrm{\&rho;}}_l\mathrm{Ls}(r_8^2-r_9^2-{\mathrm{mr}}_{10}^2)}{2};$

Thermal convection c ₉From node J ₉Flow into this resistor network; 40): with the thermal contact resistance reduction of axle is resistance R ₃₇And R ₃₈The resistor network of forming, resistance R ₃₇And R ₃₈Be connected on node J ₉And J ₁Between, resistance R ₃₇And R ₃₈Tie point be node J ₁₀,

$R_{37}=\frac{1}{2\mathrm{\&pi;}{k}_{s}L}+\frac{l_m}{2\mathrm{\&pi;}{k}_s{r}_9^2},$

$R_{38}=\frac{1}{4\mathrm{\&pi;}{k}_s{l}_b}+\frac{l_m}{2\mathrm{\&pi;}{k}_s{r}_9^2},$

L in the formula _bWide for bearing housing, l _mBe the distance of bearing center to rotor center, k _sConductivity for axle;

With the simulation on thermal convection of axle is c ₁₀:

$c_{10}={\mathrm{\&rho;}}_s{c}_s\mathrm{\&pi;}{r}_9^2(l_m+\frac{1}{2}{l}_b+\frac{1}{6}L),$

C in the formula _sBe the specific heat of axle, ρ _sBe the density of axle, thermal convection c ₁₀From node J ₁₀Flow into this resistor network.

4. the submersible electric machine with oil temperature identifying approach based on lumped parameter heat supply network network model according to claim 3; It is characterized in that: the equivalent electrical circuit that in the said step 4 each parts equivalent electrical circuit of submersible electric machine with oil is combined into whole submersible electric machine with oil according to spatial relation; And a plurality of resistance between a pair of node are carried out equivalence replace, the resistance of equivalence replacement has:

Node J ₁With node J ₁₀Between equivalent resistance R ' ₁₁₀, R ' ₁₁₀=R ₃₈Node J ₁₀With node J _{9 '}Between equivalent resistance R _{9 ' 10}, R _{9 ' 10}=R ₃₆//R ₃₇Node J _{9 '}With node J _{8 '}Between equivalent resistance R _{8 ' 9 '}, R _{8 ' 9 '}=R ₃₂//R ₃₅Node J _{8 '}With node J ₅Between equivalent resistance R _{58 '}, R _{58 '}=R ₁₈//R ₃₁Node J ₅With node J _{3 '}Between equivalent resistance R _{3 ' 5}, R _{3 ' 5}=R ₁₁//R ₁₆Node J _{3 '}With node J _{4 '}Between equivalent resistance R _{3 ' 4 '}, R _{3 ' 4 '}=R ₁₀, node J _{4 '}With node J _{2 '}Between equivalent resistance R _{2 ' 4 '}, R _{2 ' 4 '}=R ₆, node J _{2 '}With node J ₁Between equivalent resistance R _{12 '}, R _{12 '}=R ₂//R ₅

Node J ₁With node J ₇Between equivalent resistance R ' ₇₁, R ' ₇₁=R ₂₂, node J ₇With node J ₂Between equivalent resistance R ' ₂₇, R ' ₂₇=R ₃//R ₂₃, node J ₂With node J _{2 '}Between equivalent resistance R _{22 '}, R _{22 '}=R ₄,

Node J ₇With node J ₃Between equivalent resistance R ' ₃₇, R ' ₃₇=R ₇//R ₂₄, node J ₃With node J _{3 '}Between equivalent resistance R _{33 '}, R _{33 '}=R ₉

Node J ₇With node J ₆Between equivalent resistance R ' ₆₇, R ' ₆₇=R ₂₁//R ₂₀//R ₂₅, node J ₆With node J ₄Between equivalent resistance R ' ₄₆, R ' ₄₆=R ₁₃//R ₁₉, node J ₄With node J _{4 '}Between equivalent resistance R _{44 '}, R _{44 '}=R ₁₄, node J ₄With node J ₃Between equivalent resistance R ' ₃₄, R ' ₃₄=R ₈//R ₁₂, node J ₄Node J ₅Between equivalent resistance R ' ₄₅, R ' ₄₅=R ₁₅//R ₁₇, node J ₇With node J ₈Between equivalent resistance R ' ₇₈, R ' ₇₈=R ₂₆//R ₂₉, node J ₈With node J _{8 '}Between equivalent resistance R _{88 '}, R _{88 '}=R ₃₀, node J ₇With node J ₉Between equivalent resistance R ' ₇₉, R ' ₇₉=R ₂₇//R ₂₈//R ₃₃, node J ₉With node J _{9 '}Between equivalent resistance R _{99 '}, R _{99 '}=R ₃₄

The thermal value of stator winding is u in the above-mentioned equivalent circuit structure ₄, the end winding thermal value be u ₆, stator tooth thermal value be u ₃, the rotor winding thermal value be u ₈, the relational expression between the gross calorific power u of said four thermal values and submersible electric machine with oil is: u=u ₃+ u ₄+ u ₆+ u ₈

5. the submersible electric machine with oil temperature identifying approach based on lumped parameter heat supply network network model according to claim 4 is characterized in that:

Lumped parameter heat supply network network model described in the step 4 is:

θ wherein ₁Be support temperature, θ ₂Be stator yoke temperature, θ ₃Be stator tooth temperature, θ ₄Be stator winding temperature, θ ₅Be air gap temperature, θ ₆Be end winding temperature, θ ₇Be end air gap temperature, θ ₈Be rotor winding temperature, θ ₉The rotor yoke temperature, θ ₁₀Be the axle temperature degree.

6. the submersible electric machine with oil temperature identifying approach based on lumped parameter heat supply network network model according to claim 5; It is characterized in that: the method that said step 5 is found the solution lumped parameter heat supply network network model is: adopt the classical Runge-Kutta method of quadravalence to find the solution lumped parameter heat supply network network model, the primary iteration process formula of the classical Runge-Kutta method of said quadravalence is:

$\left\{\begin{array}{c}{y}_{n+1}=y_n+\frac{h}{6}(K_1+{2K}_2+{2K}_3+K_4)\\ K_1=f(x_n,y_n)\\ K_2=f(x_n+\frac{h}{2},y_n+\frac{h}{2}{K}_1)\\ K_3=f(x_n+\frac{h}{2},y_n+\frac{h}{2}{K}_2)\\ K_4=f(x_n+h,y_n+h{K}_3)\end{array}\right.,$

Y in the formula _N+1Be the functional value of the n+1 time iteration, y _nBe the functional value of the n time iteration, K ₁, K ₂, K ₃, K ₄Be respectively intermediate variable, h is an iteration step length.

7. the submersible electric machine with oil temperature identifying approach based on lumped parameter heat supply network network model according to claim 6 is characterized in that: the expression formula that obtains the temperature value of each parts of submersible electric machine with oil in the said step 5 is:

$\left[\begin{array}{c}{\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_1\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_2\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_3\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_4\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_5\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_6\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_7\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_8\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_9\\ {\stackrel{\&CenterDot;}{\mathrm{\&theta;}}}_{10}\end{array}\right]=\left[\begin{array}{cccccccc}-0.576& 0.6& 0.07& 0.04& 0& 0& 0& 0.01\\ 0.41& 0.64& 0.14& 0.08& 0.01& 0& 0& 0\\ -0.22& 0.62& -2.85& 1.87& 0.01& -2.28& 4.23& -1.37\\ -15.6& 43.8& 95.83& -123.88& 0.36& 0& 0& 0\\ 0& 0.04& 0.01& 0& -0.19& 0.123& 0& 0\\ 0.07& 0.2& 0.56& 0.39& 0.07& 15.55& -24.4& 7.93\\ 0& 0& 0& 0& 0.04& 233.62& -238.69& 89.67\\ 0& 0.006& 0& 0& 0& 3.64& -4.49& 0.85\end{array}\right]\left[\begin{array}{c}{\mathrm{\&theta;}}_1\\ {\mathrm{\&theta;}}_2\\ {\mathrm{\&theta;}}_3\\ {\mathrm{\&theta;}}_4\\ {\mathrm{\&theta;}}_5\\ {\mathrm{\&theta;}}_6\\ {\mathrm{\&theta;}}_7\\ {\mathrm{\&theta;}}_8\\ {\mathrm{\&theta;}}_9\\ {\mathrm{\&theta;}}_{10}\end{array}\right]+\left[\begin{array}{c}0\\ 0\\ 0.052\\ 19.93\\ 0.011\\ 0.077\\ 0\\ \text{0}\end{array}\right]$

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